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Sets Test - 26

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Sets Test - 26
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Let $$A = \{2,3,5,7,8,11\}$$ then which among the following is true?
    Solution
    $$2\in A$$ means 2 is element of A which is true.
  • Question 2
    1 / -0
    Which of the following statements is true
    Solution
    Element/ number three is denoted by $$3$$ and not $$\{3\}$$.
    $$\{3\}$$ denotes a subset containing element 3. Hence
    $$\{3\}\subseteq \{1,3,5\}$$
    And 
    $$3\in \{1,3,5\}$$
  • Question 3
    1 / -0
    Let $$P = \{ x | x$$ is a multiple of $$3$$ and less than $$100 $$ ,$$x$$ $$\displaystyle \in $$ $$N \}$$
    $$Q = \{ x | x$$ is a multiple of $$10$$ and less than $$100$$, $$x$$ $$\displaystyle \in$$ $$N\}$$
    Solution
    P = $$\{ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 ..., 60, ...,90, ....,99\}$$
    Q = $$\{10, 20, 30, 40, 50, 60, 70, 80, 90 \}$$
    Then looking at the multiple choices we get 
    $$P \displaystyle \cap Q$$= $$\{30, 60, 90\}$$
    $$\displaystyle \Rightarrow $$P $$\displaystyle \cap $$ Q = {x$$\displaystyle \mid $$ x is a multiple of 30 x$$\displaystyle \epsilon $$ N}
  • Question 4
    1 / -0
    If X and Y are two sets then $$\displaystyle X\cap (Y\cup X)'$$ equals:
    Solution
    We have $$\displaystyle X\cap \left ( Y\cup X \right )'=X\cap \left ( Y'\cap X' \right )$$
    = $$\displaystyle X\cap Y'\cap X'=X\cap X'\cap Y'$$
    = $$\displaystyle \phi \cap Y'= \phi $$
  • Question 5
    1 / -0
    If $$n (A) = 115, n(B) = 326, n(A - B) = 47$$, then $$\displaystyle n(A+B)$$ is equal to 
    Solution
    $$n(A-B)=47\Rightarrow n(A)-n(A\cap B) = 47\Rightarrow n(A\cap B)=115-47=68$$

    Hence $$n(A+B)=115+326-68=373$$
  • Question 6
    1 / -0
    Which of the following statements is true ?
    Solution
    $$\displaystyle 3\quad \subseteq \quad \left\{ 1,3,5 \right\} $$

    An element cannot be a subset of another set. Hence,false.

    $$\displaystyle 3\quad \in \quad \left\{ 1,3,5 \right\} $$

    $$3$$ lies in the set. Hence, true.

    $$\displaystyle \{ 3\} \quad \in \quad \left\{ 1,3,5 \right\} $$

    $$\{ 3\}$$ doesnot lie in the set. Hence, false.

    $$\displaystyle \{ 3,5\} \quad \in \quad \left\{ 1,3,5 \right\} $$

    $$\{ 3,5\}$$ doesnot lie in the set. Hence, false.

    Hence, option B.

  • Question 7
    1 / -0
    Given $$\displaystyle \xi $$ = {x : x is a natural number}
    A = {x : x is an even number x $$\displaystyle \in $$ N}
    B = {x : x is an odd number, x $$\displaystyle \in $$ N}
    Then $$\displaystyle (B\cap A)-(x-A)=....$$
    Solution
    $$A = \{2, 4, 6, 8, ........\}$$
    $$B = \{1, 3, 5, 7, .......\}$$
    $$\displaystyle B\cap A=\left \{ 2,4,6,8,..... \right \}\cap \left \{ 1,3,5,7,.... \right \}=\phi $$
    $$\displaystyle \xi -A=\left \{ 1,2,3,4,5,6,.... \right \}-\left \{ 2,4,6,8,.... \right \}=\left \{ 1,3,5,7,.... \right \}$$
    $$\displaystyle \therefore B\cap A-(\xi -A)=\phi -\left \{ 1,3,5,7,.... \right \}=\phi $$
  • Question 8
    1 / -0
    If n (A) = 120, N(B) = 250 and n (A - B) = 52, then find $$\displaystyle n(A\cup B)$$
    Solution
    $$\displaystyle n(A-B)=n(A)-n(A\cap B)$$
    $$\displaystyle \Rightarrow 52=120-n(A\cap B)$$
    $$\displaystyle \Rightarrow n(A\cap B)=120-52=68$$
    Now $$\displaystyle n(A\cup B)=n(A)+n(B)-n(A\cap B)$$
    $$\displaystyle =120+250-68$$
    $$=302$$
  • Question 9
    1 / -0
    If A and B are two disjoint sets and N is the universal set then $$\displaystyle A^{c}\cup \left [ \left ( A\cup B \right )\cap B^{c} \right ]$$ is
    Solution
    Since $$A$$ and $$B$$ are disjoint sets $$B^c\cap(A\cup B)$$ will be only $$A$$ as there is no intersection between $$A$$ and $$B$$.

    Of course $$A^c\cup A=N$$
  • Question 10
    1 / -0
    Suppose $$\displaystyle A_{1},A_{2}.....A_{30}$$ are thirty sets having 5 elements and $$\displaystyle B_{1},B_{2}....B_{n}$$ are n sets each with 3 elements. Let $$\displaystyle \bigcup_{i=1}^{30}Ai=\bigcup_{i=1}^{n}Bj=S$$ and each elements of S belongs to exactly 10 of the Ai's and exactly 9 of the Bj's. Then n is equal to 
    Solution
    Since each element of S belongs to $$10$$ $$ A_{i}s$$, there are only $$15 $$elements in S $$\left[\dfrac{30\times 5}{10}\right]$$

    Since each of the $$15$$ elements belongs to 9 $$B_j s$$ there are 135 elements in the second union.

    No. of sets $$=\dfrac{135}{3}=45$$
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